Three identical balls 1,2,3 are suspended on springs one below the other as shown.OA is a weightless thread.(a)If the thread is cut,the system starts falling.Find the acceleration of all the balls at the initial instant(b)Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread.
Asked by m.nilu
27th July 2018,
1:07 PM
Answered by Expert
Answer:
Free body diagram for all the cases, i.e., (1) at Equilibrium, (2) When the string between ball-A and support is broken,
(3) When the spring between ball-A and ball-B is broken, and (4) When the spring between ball-B and ball-C is broken are given in figure.
It is assumed that all balls are of same mass m and both springs have same force constant k.
Let x1 be the elongation of spring between A and B. Let x2 be the elongation of spring between B and C.
Let aA, aB and aC are acceleration of balls A, B and C respectively
(1) equilibrium
Let T is the tension in the string between Ball-A and support
for Ball-C :- k×x2 = m×g ..............(1)
for Ball-B :- m×g + k×x2 = k×x1 .................(2)
using eqn.(1), 2×m×g = k×x1 ...................(3)
for Ball-A :- T = m×g + k×x1 .........................(4)
using eqn.(3) T = 3×m×g ........................(5)
(2) When string between ball-A and support is broken
for Ball-A :- m×g + k×x1 = m×aA .............(4)
using eqn.(3), we can rewrite eqn.(4) as , 3×m×g = m×aA , aA = 3×g
for Ball-B :- m×g + k×x2 - k×x1 = m×aB ...................(5)
using eqn.(2), we know that LHS of eqn.(5) is zero, hence aB = 0
for Ball-C:- m×g - k×x2 = m×aC .......................(6)
using eqn.(1), we know that LHS of eqn.(6) is zero, hence aC = 0
(3) When spring between ball-A and ball-B is broken
Let T1 is the tension in the string between Ball-A and support
for Ball-A :- m×g = T1 .............(7)
m×g - T1 = m×aA .............(8)
LHS of eqn.(8) is zero because of eqn.(7), hence aA = 0
for Ball-B :- m×g + k×x2 = m×aB ...................(9)
using eqn.(1), we rewrite eqn.(9) as, 2×m×g = m×aB
Hence aB = 2×g
for Ball-C:- m×g - k×x2 = m×aC .......................(10)
using eqn.(1), we know that LHS of eqn.(6) is zero, hence aC = 0
(3) When spring between ball-B and ball-C is broken
Let T2 is the tension in the string between Ball-A and support
for Ball-A :- m×g + k×x1 = T2 .............(11)
m×g + k×x1 - T2 = m×aA .............(12)
LHS of eqn.(12) is zero because of eqn.(11), hence aA = 0
for Ball-B :- k×x1 - m×g = m×aB ...................(12)
using eqn.(3), we rewrite eqn.(9) as, m×g = m×aB
Hence aB = g (upwards)
for Ball-C:- m×g = m×aC .......................(6)
hence aC = g (downwards)
Answered by Expert
10th August 2018,
4:50 PM
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